Question

A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/hr more, the time taken for the journey would have been 1 hr 40 minutes less. Find the original speed of the car.

Answer 1

Given that the distance needed to be covered by a car is given in kilometers. Also given that if the speed of the car is a bit more than the original speed of the car, then given that the time taken by the car for the journey would be less than the original time taken by the car, if it would have maintained the original speed. Now we are asked to find the original speed of the car.

Complete step by step answer:
Converting 1 hr 40 minutes into hours, as given below:
60 minutes = 1 hour
40 minutes = 2/3 hour.
1 hr 40 minutes = $\left( {1 + \dfrac{2}{3}} \right)$hours.
$\therefore$1 hr 40 minutes = $\dfrac{5}{3}$ hours.
Let the distance of the car to be covered be $s = 400$km
Let the original speed be $v = x$ km/hr.
The original time taken in hours = $t$ hrs.
To cover the same distance by the car $s = 400$km,
Given that if the speed had been ${v'} = x + 12$ km/hr.
Then the time taken to cover the journey is = $\left( {t - \dfrac{5}{3}} \right)$ hrs.
The distance formula is given by the product of the velocity and time taken, which is given by:
$\Rightarrow s = vt$
$\Rightarrow s = xt$
In the first case, $v = x$ km/hr, and time taken is $t$ hrs.
$\Rightarrow 400 = xt$
In the second case, ${v'} = x + 12$ km/hr, and time taken is $\left( {t - \dfrac{5}{3}} \right)$ hrs.
$\Rightarrow s = v'\left( {t - \dfrac{5}{3}} \right)$, to cover the same distance of 400 km.
$\Rightarrow s = (x + 12)\left( {t - \dfrac{5}{3}} \right)$
$\Rightarrow s = xt - \dfrac{5}{3}x + 12t - 12\left( {\dfrac{5}{3}} \right)$
$\Rightarrow 400 = 400 - \dfrac{5}{3}x + 12t - 20$
$\Rightarrow \dfrac{5}{3}x - 12t + 20 = 0$
We know that $400 = xt$,
$\Rightarrow t = \dfrac{{400}}{x}$,
Substitute this value in the $\dfrac{5}{3}x - 12t + 20 = 0$ expression:
$\Rightarrow \dfrac{5}{3}x - 12\left( {\dfrac{{400}}{x}} \right) + 20 = 0$
$\Rightarrow \dfrac{{5x}}{3} - \dfrac{{4800}}{x} + 20 = 0$
Simplifying the above expression, as given below:
$\Rightarrow \dfrac{{5{x^2} - 4800 + 20x}}{{3x}} = 0$
$\Rightarrow 5{x^2} - 4800 + 20x = 0$
Taking the above expression by 5, as given below:
$\Rightarrow {x^2} + 4x - 960 = 0$
Solving the above quadratic expression gives the value of $x$, as given below:
$\Rightarrow x = \dfrac{{ - 4 \pm \sqrt {{4^2} - 4(1)( - 960)} }}{{2(1)}}$
$\Rightarrow x = \dfrac{{ - 4 \pm \sqrt {3856} }}{2}$
$\Rightarrow x = \dfrac{{ - 4 \pm 62.09}}{2}$
The value of $x$ can’t be negative, as it is the speed of a car in km/hr.
$\Rightarrow x = \dfrac{{ - 4 + 62.09}}{2}$
$\Rightarrow x = \dfrac{{58.09}}{2}$
$\therefore x = 29.04$

The original speed of the car is 29.04 km/hr.

Note: While solving this problem we should keep in mind that the distance needed to be covered by the car is the same 400 km, whether travelled with original speed and the original time, or travelled with a bit more speed and time covered is less. Hence to find out the original speed both cases are equated, as the car needs to cover the same distance.